What are the solutions to the equation (4x+1)(x-1)=4 ?

The solutions to the equation (4x+1)(x-1)=4 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8

Find the zeros of (4x+1)(x-1)=4

The zeros of (4x+1)(x-1)=4 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8

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(4 x + 1) (x - 1) = 4

x = \frac{3 + 89}{8}, x = \frac{3 - 89}{8}

Decimal Notation

x = 1.55424 \dots, x = - 0.80424 \dots

Solution steps

(4 x + 1) (x - 1) = 4

Expand

(4 x + 1) (x - 1) : 4 x^{2} - 3 x - 1

4 x^{2} - 3 x - 1 = 4

Move

4

to the left side

4 x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 (- 5) = 89

x_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 4}

x = \frac{- ( - 3 ) + 89}{2 \cdot 4} : \frac{3 + 89}{8}

x = \frac{- ( - 3 ) - 89}{2 \cdot 4} : \frac{3 - 89}{8}

The solutions to the quadratic equation are:

x = \frac{3 + 89}{8}, x = \frac{3 - 89}{8}

(a^{6} b^{24})^{\frac{1}{6}}

- 12 = - 32 + 4 x

(1.010 1^{\frac{365}{40.415}} - 1) \cdot 100

t^{2} - 6 t = - 5

(\frac{1}{25})^{\frac{3}{2}}

What are the solutions to the equation (4x+1)(x-1)=4 ?
The solutions to the equation (4x+1)(x-1)=4 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8
Find the zeros of (4x+1)(x-1)=4
The zeros of (4x+1)(x-1)=4 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8